if a=b^x,b=c^y and c= a^z prove that xyz =1 ​

Question

if a=b^x,b=c^y and c= a^z prove that xyz =1

in progress 0
Charlie 2 weeks 2021-11-16T07:50:28+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-11-16T07:51:48+00:00

    Answer:

    see explanation

    Step-by-step explanation:

    Using the rule of logarithms

    logx^{n} = nlogx

    Given

    a = b^{x} ← take the log of both sides

    loga = logb^{x} = xlogb ⇒ x = \frac{loga}{logb}

    b = c^{y} ← take the log of both sides

    logb = logc^{y} = ylogc ⇒ y = \frac{logb}{logc}

    c = a^{z} ← take the log of both sides

    logc = loga^{z} = zloga ⇒ z = \frac{logc}{loga}

    Thus

    xyz = \frac{loga}{logb} × \frac{logb}{logc} × \frac{logc}{loga} ← cancel loga, logb, logc on numerators/denominators

    Hence xyz = 1

    0
    2021-11-16T07:52:15+00:00

    Answer: xyz = 1

    Step-by-step explanation:

    a = b^x given

    introducing log into both sides

    log a = log b^x

    log a = x log b

    divide both sides by log b

    x = log a/ log b … (1)

    b = c^y given

    introducing log into both sides

    log b = log c^y

    log b = y log c

    divide both sides by log c

    y = log b/ log c … (2)

    c = a^z given

    introducing log to both sides

    log c = log a^z

    log c = z log a

    divide both sides by log a

    z = log c / log a … (3)

    Multiplying equations 1, 2 and 3 together;

    x*y *z = (log a/ log b) * (log b/ log c)

    * ( log c/ log a)

    Therefore,

    xyz = 1. proved

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )